# Haskell Quiz/Sampling/Solution Kuklewicz

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< Haskell Quiz | Sampling(Difference between revisions)

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pick :: Int -> Int -> StdGen -> [Int] | pick :: Int -> Int -> StdGen -> [Int] | ||

pick r n g | 0<=r && r<=n = pick' r n g 0 | pick r n g | 0<=r && r<=n = pick' r n g 0 | ||

− | | otherwise = | + | | otherwise = error "r must be between 0 and n" |

where pick' 0 _ _ _ = [] | where pick' 0 _ _ _ = [] | ||

pick' r n g1 i | r==n = [i..(i+n-1)] | pick' r n g1 i | r==n = [i..(i+n-1)] |

## Revision as of 13:07, 27 October 2006

This puzzle seems far too simple for the use of data structures or even monads. A single tail recursive helper function does the trick.

-- Linear solution by Chris Kuklewicz <haskell@list.mightyreason.com> -- It is important to realize you are picking from the possible -- combinations of the digits from 0 to (n-1). The probability that -- an element is chosen is (r/n). This "rolls the dice" for each element -- of the range in ascending order. module Main where import System.Random import System(getArgs) main = do [r,n] <- fmap (map read) getArgs g <- newStdGen mapM_ print (pick r n g) pick :: Int -> Int -> StdGen -> [Int] pick r n g | 0<=r && r<=n = pick' r n g 0 | otherwise = error "r must be between 0 and n" where pick' 0 _ _ _ = [] pick' r n g1 i | r==n = [i..(i+n-1)] | otherwise = let (x,g2) = randomR (1,n) g1 in if x <= r then i : (pick' (pred r) (pred n) g2 $! (succ i)) else pick' r (pred n) g2 $! (succ i)